Solutions to Systems — Question 8

Question 8 – Solution

Strategy. Factor out the supplied rotating matrix and test periodicity on the full solution, not just on its coefficients and forcing.

Step 1: Verify the proposed family directly. The product rule gives $$X_c'=AR(t)(c+t e_1)+R(t)e_1=AX_c+R(t)e_1.$$ Also $X_c(0)=c$. Thus every initial state has a member of this family.

Step 2: Prove completeness independently. Set $Z=R(-t)X$. Differentiating the inverse identity gives $(R^{-1})'=-R^{-1}A$. Consequently $$Z'=-R^{-1}AX+R^{-1}(AX+Re_1)=e_1.$$ It follows that $Z=c+t e_1$, so $\boxed{X=R(t)(c+t e_1)}$ is the entire family, valid on the real axis.

Step 3: Examine one full forcing period. Periodicity of $R$ gives $$\boxed{X(t+2\pi)-X(t)=2\pi R(t)e_1.}$$ This difference has length $2\pi$ and never vanishes, independently of $c$. In particular $X(2\pi)=X(0)+(2\pi,0)^T$. No initial state gives a $2\pi$-periodic solution. A periodic forcing can produce a response with an accumulating change rather than a periodic response.

Step 4: Classify the general constant-amplitude forcing. The same calculation gives $$\boxed{X=R(t)(c+t b),\qquad X(t+2\pi)-X(t)=2\pi R(t)b.}$$ Since $R(t)$ is invertible, a periodic solution exists exactly when $b=0$, that is, $\alpha=\beta=0$. In this case every member $X=R(t)c$ is $2\pi$-periodic, including the zero solution. If $b\ne 0$, no choice of $c$ can cancel the per-period difference. The classification concerns this specified forcing family; it does not claim that every nonzero periodic forcing precludes a periodic response.